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Inequality in four variables

To prove that

\[ \left( \frac{5a}{12}+ \frac{b}{3}+ \frac{c}{6}+ \frac{d}{12} \right) ^2 \le \frac{5a^2}{12}+ \frac{b^2}{3}+ \frac{c^2}{6}+ \frac{d^2}{12} \]

\( a,b,c,d \in \mathbb{R^+}\)

Shared by my friend Narmad Raval

Note by Megh Parikh
3 years, 1 month ago

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By Cauchy-Schwarz inequality,

\( \left( \dfrac{5}{12} + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{12} \right)\left(\dfrac{5a^2}{12} + \dfrac{b^2}{3} + \dfrac{c^2}{6} + \dfrac{d^2}{12}\right) \geq \left( \dfrac{5a}{12} + \dfrac{b}{3} + \dfrac{c}{6} + \dfrac{d}{12} \right)^2 \)

\( \Rightarrow \left(\dfrac{5a^2}{12} + \dfrac{b^2}{3} + \dfrac{c^2}{6} + \dfrac{d^2}{12}\right) \geq \left( \dfrac{5a}{12} + \dfrac{b}{3} + \dfrac{c}{6} + \dfrac{d}{12} \right)^2 \) Siddhartha Srivastava · 3 years, 1 month ago

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Expand To Get

\(\frac {5}{36} (a-b)^2 + \frac {5}{72} (a-c)^2 + \frac {5}{144} (a-d)^2 + \frac {1}{18} (b-c)^2 + \frac {1}{36} (b-d)^2 + \frac {1}{72} (c-d)^2 \geq 0\) Anant Chhajwani · 3 years, 1 month ago

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@Anant Chhajwani Thanks .

I dont know how I could miss that. Megh Parikh · 3 years, 1 month ago

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